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How does a zero g aircraft work?
In upward motion, drag and gravity is in the same direction (both in downward direction). So in this situation G is negative or less than 1 or something else?
In downward motion with thrust greater than drag (so we have acceleration), whats will happen for G?

I believe you'll get a better answer on physics.stackexchange.

I think the question is insufficiently constrained to clearly ask a single answerable question.  Last two paragraphs of my answer (before footnotes) should explain why.

A zero G flight is called like that because the people inside the aircraft do not feel any force relative to the aircraft during the zero-gravity maneuver, meaning there is no acceleration (or G force) relative to the aircraft and people therefore experience weightlessness.
However, that does not mean that the aircraft experiences zero G with respect to the Earth. The aircraft is still accelerated by the forces acting on it (like gravity).
The zero-gravity maneuver is achieved by flying on a specific parabola. Imagine throwing some object upwards and sideways. The object will move on a parabola because of gravity with a constant acceleration of 1 G in the downwards direction (ignoring drag). An aircraft can be intentionally flown on such a parabola, which then causes the zero G effect inside the aircraft. Aerodynamic forces (lift and drag) and thrust are not negligible here, so the aircraft must be carefully flown on a parabolic path with exactly 1 G acceleration in the downwards direction. This acceleration is maintained during the entire weightlessness period (both upwards and downwards part):

What is a parabolic maneuver?
The parabolic maneuver or ellipse arc during a Zero G flight is divided into three stages: the parabola pull-up, the parabola, and the parabola pull-out.
Pull-up
The pilot lifts the nose of the A310 Zero G airbus upward from horizontal flight to an angle of 50 degrees.
Passengers experience a pull of 1.8 times that of gravity on Earth. This is called hypergravity.
The parabola pull-up or “nose-up” lasts around 20 seconds.
Parabola
The parabola starts with injection: as the aircraft travels upward the pilots reduce engine speed and the Zero G aircraft follows a ballistic trajectory.
Weightlessness begins when the aircraft enters a parabola during which it is in free fall for 22 seconds.
Pull-out
The nose of the plane is tilted back downward 42 degrees, before the pilots gradually level off the aircraft as they increase engine speed.
Passengers again experience a pull of 1.8 times that of gravity on Earth.
After 20 seconds, the A310 Zero G returns to a horizontal trajectory.

(AirZeroG - How zero-gravity parabolic flights work)
Flying on such a parabola is not easy. The A310 by AirZeroG is flown by 3 pilots at the same time to carefully stay on the parabola:

Parabolic maneuvres piloting technique
Unlike commercial flights, the Airbus A310 Zero G is simultaneously piloted by three members of the flying crew during the parabolic manuevres.

One pilot controls the pitching (nose-up and nose-down angle).
A second pilot controls the roll movement (to keep the wings horizontal).
A third pilot, sitting behind them, controls the engine speed (he or she also monitors the flight parameters: warnings, temperatures and pressure).

Together, all three pilots maintain a near-zero acceleration level in the three axes to guarantee zero-gravity precision ± 0.02 g
Unlike in a standard aircraft, the pitching and roll control commands are dissociated. The two control column functions are dissociated by the use of an additional control column on one side and two small cables hung from the conventional control column.

(AirZeroG - How zero-gravity parabolic flights work)

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Key point: the force of gravity "acts from within"-- it's an internal force, not an external force.  Gravity exerts an equal force per unit mass on every molecule of the aircraft structure and the aircraft contents (including the bodies of the crew and passengers).  Thus if the only force acting on the aircraft and content is gravity, every molecule of the whole assembly accelerates together in a way that creates no stresses or strains of any kind on the aircraft structure and contents.  Unrestrained bodies will float freely.  This is "weightlessness".
The same is true if the net force acting on the aircraft is exactly equal to the force of gravity.  In other words, if all the aerodynamic and thrust forces cancel each other out.  To a first approximation-- specifically, assuming that the thrust line from the engines acts exactly parallel to the flight path-- this is the case whenever lift is zero, and thrust is regulated to be exactly equal to drag.  Maintaining this condition with great precision is what the crew is doing with the interesting "split" control system described in another answer to this question.
The three-dimensional "G-loading"1 experienced by the aircraft is the vector sum of all the aerodynamic and thrust forces, divided by mass.  When this is equal to zero, the only force acting on the aircraft and contents is the force of gravity, and the aircraft and contents experience "weightlessness", or 0-G flight.

In upward motion, drag and gravity is in the same direction (both in
downward direction). So in this situation G is negative or less than 1
or something else?

I'm assuming you are asking specifically about the upward portion of a 0-G trajectory.  I hope this answer has helped make clear that during the actual 0-G portion of the 0-G trajectory, where the G-loading is zero, lift is zero and thrust exactly equals drag. The aircraft is decelerating (losing airspeed) but not due to drag-- the deceleration is entirely due to the component of gravity that is exacting parallel to the instantaneous direction of the aircraft.  However, another answer has made clear that is preceded by an earlier phase where the flight path is curving upward and the G-loading is thus greater than 1.  To a good approximation, the total G-loading in this earlier phase (given as 1.8 Gs in the other answer) is entirely due to the extra lift force generated by the wings to curve the flight path upwards.

In downward motion with thrust greater than drag (so we have
acceleration), what will happen for G

In the portion of a 0-G trajectory where the aircraft is descending, thrust is never greater than drag, so it's hard to fit this part of your question into the context of the rest of your question.  In the actual 0-G portion of descending portion of the 0-G trajectory, thrust is exactly to drag, and lift is zero.  Airspeed is increasing not because thrust is greater than drag, but rather because the net force acting along the instantaneous direction of the flight path at any given moment, including the component of gravity acting in that direction, is greater than zero.  In the subsequent recovery phase of the 0-G trajectory, lift is actually greater than the weight of the aircraft and contents, so the G-loading is greater than 1, as the flight path curves upward back toward horizontal.
But you seem to have shifted gears away from the 0-G situation to ask about a situation where the aircraft is descending and thrust is greater than drag.  In any case where thrust is greater than drag, regardless of whether the aircraft is ascending or descending, the aircraft and occupants will experience a forward G-loading component.  I.e. the crew and passengers will "feel" a force component pushing forward against their bodies from the back of their seats.  This is true regardless of whether the net 3-dimensional G-loading is 1-G earthwards (think of a sustained steady climb at constant airspeed with the aircraft in a nose-high pitch attitude), or otherwise (consider a vertical dive with increasing airspeed), but it could never happen when the net 3-dimensional G-loading was exactly zero.  So this condition would have nothing to do with 0-G flight.
As to the specific part of question that asks "what will happen for G", the answer can only be "it depends-- you haven't given us enough information to answer."  You've told us that thrust is greater than drag, so we know there's a positive G-load in the fore-and-aft sense-- the back of the seats are pressing forward against the bodies of the crew and passengers.  But just telling us that the flight path is "descending" isn't enough to know whether the lift vector from the wings is greater than 1 G positive, or less than 1 G positive, or zero, or negative.  (Some of these cases would only be consistent with some rather aerobatic maneuvering!)  So the magnitude and direction of the "up and down" component (in the aircraft's own reference frame) of the G-loading vector remains unknown, given the information contained in the question.
If we assume the aircraft is descending along a constant linear trajectory while accelerating2, then the "up and down" component (in the aircraft's own reference frame) of the G-loading vector must equal the cosine of the descent angle.   In such a case, at steep descent angles, the total three-dimensional G-loading could be less than 1 G, even though we've specified that the fore-and-aft component of the G-loading is not zero.  The question doesn't contain enough information to say more than that.  Of course, the magnitude of the fore-and-aft component of the G-loading vector is a factor too-- if thrust is enough greater than the drag, this component of the G-loading alone could be greater than 1 G, in which case the total three-dimensional G-loading certainly would be as well!
Footnotes:

Re "three-dimensional" -- sometimes G-loading is used in a one-dimensional sense, to simply mean the acceleration component acting in the upward direction in the aircraft's own reference frame.  This is what a traditional mechanical panel-mounted G-meter measures.  In many situations, this is only one component of the total three-dimensional G-loading.

Note that it would not be necessary for thrust to be greater than drag for an aircraft to accelerate while descending with a linear trajectory.  A roller coaster, with zero thrust, can accelerate along a linearly descending track, up to a certain speed where drag equals to the "forward" component of the gravity vector.

Related questions -- https://aviation.stackexchange.com/questions/48115/why-do-we-feel-gravity-on-a-plane , https://aviation.stackexchange.com/questions/48136/why-do-zero-gravity-flights-not-stall , https://aviation.stackexchange.com/questions/80923/how-did-pilots-maintain-the-vomit-comets-parabolic-trajectory

Also related: https://aviation.stackexchange.com/questions/2890/why-cant-an-airplane-stall-at-zero-g?rq=1

It seems you are having a discussion with yourself. If you want to expand your own answer based on these comments, please do. I'll move this to chat.

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@quiet Flyer, "acceleration component acting in the upward direction in the aircraft's own reference frame"?  Acceleration, by definition, is the rate of change of velocity, which is the rate of change of position. Clearly, a moments thought makes it clear that in the frame of reference of the aircraft itself, the G-meter is not moving. A G-meter measures the inertial acceleration of the G-meter, i.e., the acceleration that would be measured in an inertial, (zeroz-G, free falling frame.

(answered in chat)


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In a banked turn, why does the center of gravity, rather than the center of lift, follow the path of a circular arc?
I'm a post-solo student pilot who sometimes gets hung up on fundamental points of aerodynamics.
"As long as the aircraft is banked, the side force is a constant, unopposed force on the aircraft. The resulting motion of the center of gravity of the aircraft is a circular arc." (https://www.grc.nasa.gov/www/k-12/airplane/turns.html)
I'm having a difficult time understanding why the center of gravity is translated along this circular arc, when the turning force that pulls the airplane through the turn is acting upon the airplane's center of lift, aft of the CG. It makes intuitive sense to me why the airplane rotates about the CG on all of its axes, but all translation / movement being located at the CG is more difficult for me to visualize.
My intuition is that the turning force acting aft of the CG would cause the plane to rotate (pitch) about the CG, rather than turn. I know that in straight and level flight, this tendency is counteracted by the downward force exerted on the horizontal stabilizer. I'm assuming this point is also relevant in answering my question, but it still strikes me as counterintuitive.
The PHAK describes the relationship between CG, CL, and tail as a "lever," balanced on the CL, with the weight of the CG and the downforce of the tail on either side. If the force that turns the airplane stems from the balance point of that lever, I would expect that to be the point that follows the direction of the turn, rather than the weight on one side of it!
Thanks for any insight, and I hope that I was clear in my phrasing of this question.
Cheers.

I'm having a difficult time understanding why the center of gravity is
translated along this circular arc, when the turning force that pulls
the airplane through the turn is acting upon the airplane's center of
lift, aft of the CG. It makes intuitive sense to me why the airplane
rotates about the CG on all of its axes, but all translation /
movement being located at the CG is more difficult for me to
visualize.

The truth is that in turning flight, every point on the aircraft is following a circular arc.  There's no reason to single out the CG.  Note however that the circular arc followed by each point on the aircraft is not the same.  For example, the inboard wingtip follows an arc of smaller radius than the outboard wingtip does.  This is important to understanding the "overbanking" tendency that some aircraft experience-- since the outboard wingtip covers more distance (circumference) in a given unit time than the inboard wingtip, the outboard wingtip experiences a higher airspeed than the inboard wingtip, and so tends to generate more lift than the inboard wingtip.
We can think of the instantaneous linear velocity (speed and direction) of any point on the aircraft as the result of the aircraft's instantaneous linear velocity at the CG, acting tangent to the circumference of the turn, plus an additional linear velocity component due to the aircraft's rotation about the CG.  But guess what -- we're just using the CG as a handy reference point here.  It's equally valid to pick any other point on the aircraft as our reference point.  For example, we could pick a reference point on the inboard wingtip, measure its instantaneous linear velocity, and then calculate the instantaneous linear velocity of every other point on the aircraft by treating the problem as if the aircraft were rotating around the inboard wingtip, rather than rotating about the CG.  Either way of treating the problem would be equally valid, and either way would give the same result for the instantaneous linear velocity of any point on the aircraft.  For example, we'll still find that the outboard wingtip is moving through the air faster than the inboard wingtip, and we'll still get the same number for the difference in instantaneous linear velocity between the inboard wingtip and the outboard wingtip.  There's no rule that says we must pick the CG of the aircraft as our reference point or "pivot point" for these sorts of calculations.  It's not really valid to say that a moving, rotating body rotates about its CG and not about some other pivot point.  The choice of a "pivot point" is arbitrary, and need not be at the CG.
If the "center of lift" (which you seem to be defining as where the upward lift from the wing acts, without regard to where the upward or downward lift from the tail acts) were somewhere other than at the CG, you would be perfectly valid in choosing that center of lift as the reference point for the sort of calculation we've been describing here.  But you could equally well choose the CG for your reference point.  It's arbitrary.
However, your question does show a misconception that the "center of lift" of the entire aircraft is somewhere other than at the CG, at least in the fore-and-aft sense. Remember, the horizontal tail can generate either downward lift or upward lift.  As far as the static balance of pitch torques is concerned, downward lift generated by the tail has the same effect as upward lift generated by a canard wing, and the situation can be analyzed as if the aircraft had an uplifting canard wing rather than a downlifting tail.  Either way, the "effective center of lift" of the wing-tail system, i.e. of the entire aircraft, must be co-located with the CG, at least in the fore-and-aft sense, or a net pitch torque would be created, and the aircraft would be experiencing an increase or decrease in the rate of pitch rotation.
(Disclaimer-- we're assuming here that thrust and drag are equal and opposite and act in line with each other.  An engine mounted high above the aircraft on a tall pylon, for example, would tend to generate nose-down pitch torque that could be offset by moving the center of lift of the entire aircraft ahead of the CG.)
There's one other significant twist though.  If we are interested in calculating the balance of pitch torques on a rotating body, rather than simply calculating instantaneous linear velocities at various points on the body as described above, then if we choose any point other than the CG as our reference point, then we have to introduce a fictitious pseudoforce called "centrifugal force" which acts at the CG and is generated by the CG's curving trajectory around the center of the turn.  Choosing the CG as our reference point avoids the need for introducing this pseudoforce, so "centrifugal force" need not enter our calculations.

Thanks for the thorough reply! Yes, I was defining CL as the upward force of the wings, separate from the downward force of the tail (PHAK, Figure 5-23). The idea that a CL not colocated with the CG would cause a net pitch torque is exactly what I was referring to in my 4th paragraph. The point that an "effective center of lift" is colocated with the CG I think gets to the heart of my question and answers it.

Well, are you forgetting to use the tail?
The "center of gravity" model can take some time to fully fathom, but it works well to describe rotation on all three axes.
Let's try pitch.  Since CG is ahead of wing CP (center of pressure), CP will try to pitch the nose down.  So how do we "rotate" to take off?  More up elevator!.
Flying a gentle full circle the plane actually does one rotation around the yaw axis with some help of the rudder.  Steeper turns use more elevator.
We can see that the elevator and rudder provide the torque (+ or minus) that controls rotation in pitch and yaw.  Rotation on the roll axis is with ... the ailerons.
In a turn the tail coordinates the rate of rotation with the change in flight path (the "turn") caused by the horizontal component of the lift vector (the constant, unopposed side force).  The result is the nose of the plane follows the curvature of the circle as much as possible, reducing drag.
It helps to keep one a bit more "on the ball".

The centre of gravity is the reference point of the rigid body. The six degrees of freedom are modeled as three linear translations of the CoG, and three rotations around the CoG.
You’re right in that the centre of lift could be taken as the reference point as well, or any other point for that matter.  It’s just that rotations of a body suspended in air take place around the CoG, and it is the movement of the rigid body that defines the reference point.

The airplane follows an arc in the turn because of the vertical fin.  When you bank, it introduces a lateral vector to the wing's lift force.  The lateral vector makes it displace sideways as it's moving forward.  Without the fin to provide a weathervaning effect, it would simply continue more or less pointed on its original heading, but on a slewed track, so that the airflow is striking the body from ahead and to one side.
Think of a helicopter hover taxiing, moving forward very slowly.  You apply left cyclic to tilt the rotor disc a bit as you move forward.  The machine continues pointing where it was pointing, but starts to move laterally because the rotor's lift vector was tilted.  If it was pointing North, it would continue pointing North, but actually proceed in a straight line Northwest.
It won't actually start to arc around with a heading change until the lateral flow hitting the tiny little fin at the back starts to point it if you're going fast enough, or you move the tail yourself with some pedal input. When you bank the airplane, the same forces are at work.
On the plane, the offset flow causes the fin (the fixed fin, not the rudder) to generate a local lateral lift force, a weathervane effect, realigning the body into the flow.  The slewing or side slipping action followed almost immediately by a realignment by the fin results in a turning arc.
It's desirable to have a slight lag in this activity, because this initial side slipping action is required for dihedral effect to work, to provide lateral stability.  Dihedral effect requires that a little bit of lateral sideslip be allowed to develop when a bank is induced, to create the differential lift that tends to restore the plane back to level flight, before the fin starts to do its realignment function. Fin sizing has to take this into account; too big and the plane starts to align its yaw axis to the flow instantly when a bank occurs and you end up with a spiral tendency, and too small, the plane slithers and slides around.
Note that it's the fixed fin, not the rudder doing this.  If you have your feet on the floor, the rudder just trails in the airflow, not doing anything.  If the plane has no adverse yaw, you can leave your feet on the floor while maneuvering and the plane will maintain coordinated flight because the fin itself is able to continuously weathervane the body into the flow.
The C of G is simply the point at which any rotational movements are centered.  Pitch forces come into it mostly to compensate for the loss of the purely vertical component of lift that was reduced when the bank was induced, to keep from descending.  Elevator inputs change the trim force balance to increase the AOA that will be sought by the plane's static pitch stability forces.
Since increasing AOA increases the lift vector perpendicular to the body, the increase raises both the vertical and lateral lift forces, increasing both the lift required to keep from descending, and the turn rate resulting from the lateral lift vector. So turn rate is influenced by both bank angle and elevator input, because both factors influence the lateral component that wants to displace the airplane sideways, with the pitch factor only a minor part of it until the bank angle gets steep. As the bank steepens, more and more of the total turn rate is coming from pitch rate from the elevator (take it to the extreme and imagine a turn at near 90 degrees bank - nearly all the change in heading is from pitch since you are basically flying a loop laid flat).
While all that is going on, the fin just sits there keeping the body aligned into the flow, with the rudder chipping in to help out via your feet to compensate for adverse yaw or bumps, as required.

Draw a big circle to represent the path of the CG during the turn, as seen from above, and then draw the airplane on that line.  Where is the point of tangency between the longitudinal axis of the aircraft, and the curving path of the turn?  At the CG?  Well aft of the CG, due to the "weathervane" effect of the fin?  In the latter case, the forward parts of the aircraft (including the wings) are feeling a flow component toward the outside of the turn-- and dihedral is therefore creating a rolling-out torque-- even in a steady-state turn turn.

The point of tangency might even be far behind the aircraft itself-- a slipping turn-- or far ahead of the aircraft itself-- a skidding turn.  Anyway that's one way to think about the combined effects of slip/skid, and rotation, and what direction of airflow the fin (and other parts of the aircraft) will "feel" (and therefore what yaw torque the fin will create) in each case -- but I'm not sure the original question was about what the fin is doing in a turn!

My post is to make clear that the turn forms an arc because of the fin and the C of G isn't that important other than forming the axis about which pitch and yaw occurs as the center of mass, and without the fin the turn would be more of a slewing motion.  This seems to be poorly understood, as is the fact that the fin needs to small enough to be insensitive to initial side slip for dihedral effect to work.  In a sustained turn with no rudder input you will normally have a  small amount of sideslip present continuously, depending on fin sizing, since it has to be operating at some positive AOA.

"Without the fin to provide a weathervaning effect, it would simply continue more or less pointed on its original heading, but on a slewed track, so that the airflow is striking the body from ahead and to one side." I appreciate this comment a lot. This was actually the original question I had that led me to the question I ultimately ended up posting. I was wondering why a horizontal component of lift causes a heading change and not just a slewed track, as you say, which led me to the NASA quote I included in my original post, which in turn led me to the question I asked here.

When you add floats on a plane, the surface area of the floats is mostly ahead of the CG and they reduce yaw stability.  I used to fly a Feet Canuck on floats that had no additional compensating ventral fin added.  When you banked it with aileron, it had massive adverse yaw like a glider's because of the minimal weathervaning tendency.  If a bump put it into a bank, it would slip into the low wing and the nose would only slowly come around.  You had to work your feet constantly. Cessna helped the yaw stability of the 180 on floats without additional fin area by putting centering springs in...

...the rudder control circuit. The springs would tend to fix the rudder at neutral except when you overpowered the springs pushing with your feet, and this tended to help out the fin in providing passive weathervaning, having an effect similar to adding an extra stabilizing fin. Anyway, yes it's the fin pulling the nose around, in response to being blown on from the side from the initial side slip tendency, that results in the center of mass moving in an arc, in addition to whatever effect you are getting from pitch change at a given bank angle.

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If i have equal lift and drag, do they cancel each other out?
I am working on a throw glider and was wondering, How much lift would be too much lift, as in it causes the glider to stall (maybe thrust that causes this, but still). also, if I create a body that causes a large amount of drag, And then use an airfoil that causes an equal amount of lift, would they cancel each other and make the plane fly like a spear, or would one still be greater than the other? (not sure if I'm explaining correctly).

They do not. Lift and Drag do not directly oppose each other.
The opposite of Lift is Weight (gravity). The opposite of Drag is Thrust.
The performance of the craft will depend on how these two forces are balanced against their opposites.

That makes sense. thank you for your answer.

Actually, for any given glider with an L/D ratio of 1:1, if you added enough weight it would fly fast, not slow, and likewise the stall speed would be high, not low.

@quietflyer The OP has been edited, original language said "high lift...high drag" rather than 1:1. Editing now.

Yeah it's all a little tricky, the question does sort of imply that lift might be high in relation to weight but if we know it's a glider and we know L and D are equal then we know that Lift is about .7 weight.  "High lift" really likely means "high lift coefficient" which does have implications for airspeed, but it 's not stated outright.  All in all the question sort of blurs the concepts of "Lift" and "Lift Coefficient" and likewise for drag.

If Lift and Drag are equal and Thrust is zero, this simply means that -- at least in the case of steady-state flight at a constant airspeed in still air -- the glide angle will be 45 degrees.
For more, see these related ASE answers--
Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag?
Why is the L/D ratio numerically equal to the glide ratio? (all answers)
If you truly want the glider to "fly like a spear" -- i.e. to follow a completely ballistic trajectory-- then lift must be zero.  But I doubt that this is actually your goal.  Though especially in the world of paper airplanes, some models with rather dart-like properties have done rather well in distance competitions held on level ground (or the indoor equivalent).

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What means when center of pressure is out of physical dimension of wing?
Center of pressure(CoP) is point where total force vector acting at. If you hold wing at this point, wing will not rotate(moment=0). Example hanging point at hanglider.
I see at some diagrams for very low and negative AoA-s that center of pressure is out of wing physical dimensions. At diagram below  for AoA 0.5°, -3°, 4°
What does it mean in reality, no place where can I hold wing without rotation?
Or if I extend wing with some stick and hold at this position, wing will not rotate?
At first I will tell that for every AoA, must be place on wing where moment is zero.But obviusly I am wrong. Hard to imagine to me how pressure distribution around wing must be to shift CoP so much out.
That also mean if you dont know what you are doing, you can design so bed hanglider wing that will shift CoP out wing dimension for some AoAs. This is dangerous situation, theoretically speaking.

That's why the modern way to analyze the situation is treat the lift force as acting at the wing's aerodynamic center, and also include a pitching moment term.  Still, the question appears to be answerable.

"My logic tell me that for every AoA, must be place on wing where moment is zero, it can not be out of wing dimensions" is wrong. Surely, it's not out of the realm of imagination that, at some incidence angle, no matter where you place your finger on the airfoil, the airfoil would always rotate? In that scenario, the CoP is out of the dimension.

@JZYL So math again showing absolute truth. That mean you can design bed hanglider wing that will not have correct hangpoint position for some AoAs? Hmm this is scary.

@JurgenM No, that's why you have tail or you use reflex airfoil with elevon, which guarantee feasible trim point at least until buffet/stall.

@JZYL Yes but in theory you can make so bed wing  that will shift CoP out of wing.I talk theoretically

@JurgenM you could design a hang glider like that, but you probably won't have many customers for it.

@JurgenM Of course you can. Even if the CoP is within the wing, it may still be unstable and totally unflyable. What's your point?

(added another tag because-- once you start asking about where you can hold the wing without a rotation tendency, or how a hang glider pilot should be connected to the wing, this is not just pure aerodynamics)

If you want to be able to trim a hang glider to have zero pitch trim force (pilot doesn't need to push or pull on control bar) with the airfoil shown above, at 0.5 degrees angle-of-attack, assuming drag on the pilot's body is negligible, the "hang point" would have to be vertically in line with the circle on the diagram that is just aft of the trailing edge of the wing, labelled "0.5 degrees".  Connecting the hang strap right at the circle would accomplish this, but that is not only possible configuration-- it could also connect above, or below, the circle.
If drag on the pilot's body is not negligible-- so that when the pilot lets go of the control bar, the hang strap angles backwards rather than running straight up and down-- then the extended line of the hang strap will have to pass through the circle on the diagram that is just aft of the trailing edge of the wing, labelled "0.5 degrees".    Connecting the hang strap right at the circle would accomplish this, but that is not only possible configuration.  It could be attached below the circle and slightly aft of it, or above the circle and slightly forward of it.  The key point is that the extended line of the hang strap must pass through that circle.
(Note that we're only talking about trim, not stability.  Assume for simplicity that we have a straight, non-swept, constant-chord wing.  Such a wing would not be stable-- the angle-of-attack would not tend to stay constant-- at least with a conventional airfoil cambered in the usual direction-- but we can still discuss the trim characteristics at any given angle-of-attack, where the forces and torques are at least momentarily in balance.  Hang glider stability dynamics have been discussed elsewhere on ASE, and are fertile ground for additional future questions.)
Now imagine that we flip the wing upside down, so that we have an airfoil with "negative camber", and wish to fly at a 3-degree angle-of-attack.  Now to trim the glider for hands-off flight, the extended line of the hang strap will have to pass through the circle labelled "-3" on the diagram.  Again, this can be accomplished by attaching the hang strap right at the "-3" circle, but this is not the only possible configuration.
It seems somewhat counter-intuitive that the center-of-pressure could lay outside the physical dimensions of the airfoil, but it is possible.  The reason that this happens is that any non-symmetrical airfoil generates a pitching moment.  If the airfoil is cambered in the normal direction, the pitching moment acts in the nose-down direction.  Even at the zero-lift angle-of-attack, the airfoil generates a nose-down pitch torque.  The modern way to analyze the forces and torques generated by an airfoil is to treat the Lift and Drag forces as acting at a Center of Pressure which is defined to be fixed in location, at the quarter-chord point of the wing.  Any nose-up or nose-down aerodynamic pitch torque that would exist if the wing were fixed on a pivot at this Center of Pressure, is expressed as a Pitching Moment Coefficient.

You can extend with stick to circle 0.5 and put hang strap here and when wing increase AoA it will stall and kill yourself.

@JurgenM-- see my comments in answer about hang glider stability-- despite your comments, hang gliders do indeed have cambered airfoils, yet are indeed stable in pitch. I suspect that both you and I know the reasons that this is possible, but that's a whole 'nother topic entirely.


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What does it mean when an aircraft is statically stable but dynamically unstable?
I am reading this book: "Flight Stability and Automatic Control", second edition, Dr. Robert C. Nelson. In the chapter 2, page 42, the author wrote:

Note that the vehicle can be statically stable but dynamically unstable. Static stability, therefore, does not guarantee dynamic stability. However, for the vehicle to dynamically stable it must be statically stable.

I understand static stability, but I am not sure I understand dynamic stability. Can someone help me understand the above texts and give me an example of this case?

If an aircraft is statically stable, it will always return to equilibrium after a disturbance. But what happens after can either show instability or stability. This is where the dynamic stability comes in.
You can think of an aircraft at equilibrium at a particular speed, altitude and angle of attack and it is suddenly faced with a disturbance which changes its speed, altitude and angle of attack. If the aircraft has static stability, it will immediately seek its equilibrium state. If the said aircraft is also dynamically stable, the amplitude of its motion will reduce with time. This is called subsidence. One of the major factors that affect dynamic stability is the amount of damping in the system. From now on we will consider the aircraft already is statically stable. When there is enough damping, an aircraft will slowly in time, reduce its amplitude until the amplitude goes to zero. Here the aircraft is said to be dynamically stable. If there is less damping the oscillations increases with time and the amplitude of motion also increases. This is called divergent oscillation. In this situation, the aircraft is said to be statically stable but dynamically unstable.
The aircraft has less dynamic stability at high altitudes where aerodynamic damping is lower. A pilot can also reduce the dynamic stability of the aircraft. If his/ her inputs are close to the natural frequency of the aircraft it can add energy to the system and the divergence increases. This is called Pilot induced oscillation (PIO).

Statically and dynamically stable.

Statically stable but dynamically unstable.
It is important to keep in mind that an aircraft, while it can be statically stable and dynamically unstable, it cannot be the other way around. That is, an aircraft can never be statically unstable and dynamically stable.

Possible improvement-- first sentence-- change "it will always return to equilibrium" to "it will always seek to return to equilibrium"... if it has truly returned to equilibrium then it will stay there till the next disturbance--

Would it be possible for an aircraft to settle into an oscillation which reaches a certain amplitude and then remains there?  Would that be classified as dynamically stable or unstable?

@supercat not for control theory, however all material components have maximum deflections etc, and once those are reached the control system cannot react further. This might mean that a wing bends in such a shape that the dynamic stability is removed. Another reason might be that you have a secondary control system acting "over" the basic system: which adds stability but works at much lower frequency.
But those effects are ignored when discussing stability of an aircraft: you just don't want these to happen or depend on them.

@paul23: In a system of linear differential equations, a system will either be overdamped, underdamped, or critically damped; even when flying within normal parameters, however, an airplane's behavior won't be perfectly modeled by a system of linear differential equation.  Many real-world non-linear systems (including most purpose-designed oscillators) can oscillate with closed-loop gains that are greater than 1 for part of each cycle and less than one for other parts.

@supercat If the aircraft reaches a disturbance and stays at the point where the disturbance threw it to, the aircraft is said to be both statically and dynamically neutral. Airbus aircraft control laws are actually designed this way. It behaves as if it has neutral stability. Within the protection envelope if you make an input the aircraft remains there until the pilot brings in a change.

I thought I have misunderstood static stability the whole time: static stability only ensures the system has ability to go back equilibrium, this does not gurantee system will stay there. I have always thought static stability means the system will stay at equilibrium state after being disturbed after long time enough.

+1 for the plots, made the concept immediately obvious to me in a couple seconds given the background knowledge of stability I already had.

I think there's a typo: I presume neutral frequency should read natural frequency?

Static stability means that a deviation from a trimmed state produces forces which return the system to this trimmed state.
If these forces produce an overshoot which increases over time, such that the system oscillates around this trim point with increasing amplitude, the system is dynamically unstable. The long period oscillation (phygoid) of gliders is often unstable because their L/D is high enough to push them into unstable territory.
Dynamic stability means that the oscillations die down over time. Without static stability the system would simply leave the trimmed state without any tendency to return. This tendency to return is the prerequisite for the oscillation, whether stable or unstable.

+1 for the reference to L/D, I've often wondered whether when cloud-flying in a glider especially without artificial horizon, it would be helpful to lower the gear and possibly also crack the spoilers open a bit?  Maybe make some sort of block or other mechanical aid to hold the spoilers slightly open without constantly occupying the pilot's left hand?  (Just a theoretical question for me at present but I know in some other parts of the world cloud-flying in gliders is still somewhat routine...)

That'll just make more noise as you become disoriented and go into a spiral dive past Vne, which is guaranteed if you don't have gyro instruments, and likely eventually even if u had an electric turn coordinator.If the glider has speed limiting dive brakes/spoilers, meaning the plane is incapable of exceeding Vne even straight down, and most modern glider do, your best bet is to just dump full spoilers and let it go and wait to come out of the bottom. If you try to maintain controlled flight in cloud for any length of time with spoilers in or partly out, you will come out the bottom in pieces.

Well, I don't know about all that, flying w/ turn rate indicator as sole gyro instrument has long been common practice for many decades (like since the 1940's) in gliders in some countries in Europe and the UK.

@JohnK: I met an old pilot who claimed he could maintain controlled flight in clouds without a gyro, but required extreme concentration. I wouldn't want to test it, but I can see how it might be possible. Mathematically, it's integrating over the bank indicator (otherwise known as the turn coordination indicator) and the compass.

@Joshua You have to have some kind of reference to tell your eyes you're turning, because the inner ear is useless for that purpose. If you sustain a right turn say for a full circle, and the cochleal fluid stabilizes with the rotation, when you roll out you get an intense sensation of turning the other way and almost all pilots will unconsciously start to roll back into the turn. It's supposedly possible using the compass to tell you you are turning, but you would have to be very good at anticipating and accounting for turning errors, and if there was any turbulence you'd be screwed.

@quietflyer I'm having trouble believing intentional, sustained flight in IMC is an accepted practice in Europe or anywhere else that pilots are expect to be prudent and sane (Russia, well who knows lol).  A turn gyro in a glider is there for the same reason you might have one in a puddle jumper VFR-only airplane; to get you out of a cloud in one piece if you stumble into it somehow, as soon as you can.  Sustained flight on "partial panel" in cloud, on purpose, especially in an aircraft neutrally stable in roll like a sailplane, would be crazy.

@Joshua Yes, cloud flying without gyro is possible, but only in "dead" clouds without much air movement in them. If you try this in an active cumulonimbus, JohnK is right with his advice.

the example for static stability: "a ball sitting in a pit" gave me the wrong idea of static stability the whole time in my life. I thought static stability means the system will stay at equilibrium state after long time enough. This "a ball sitting in a pit" example also gives the idea about dynamic stability but people always show it when they want to talk about static stability

@JohnK -- no, partial-panel cloud flying in gliders has long been an accepted practice in some parts of the world, especially in Europe and England.
Here's a traning document from an English gliding club -- https://www.dartmoorgliding.co.uk/Instrument_Flying_course_notes.pdf -- it has been updated to mention GPS, but the first 15 pages are all about how to control the glider using only turn rate indicator, ASI, and conventional magnetic compass.  I've accumulated a library of over a dozen books on gliding from the 60's and 70's and most of them reference partial-panel cloud flying.

@JohnK -- the only one I have handy right now is "New Soaring Pilot" by Ann and Lorne Welch and Frank Irving,3rd edition, 1977.  Chapter 10, pp 120-139, is entitled "Instrument Flying", and is entirely about cloud flying.  On p.136 there is a sentence beginning "Although blind flying is perfectly practical on a turn-and-slip indicator, it is made easier of the glider is equipped with an artificial horizon..."  All preceding pages assume there is no artificial horizon, and there are plenty of references to flying in cumulus clouds, not just smooth mountain wave clouds.

@ JohnK -- a friend of mind flew gliders in Holland and elsewhere in Europe in the (60's? 70's?) and accumulated many hours of soaring in cloud, w/o artificial horizon.  I also know someone who, back in the '70s when the floor of controlled airspace may have been higher, set a state altitude record in a midwest flatland state in the US by thermalling up in a towering thermal cloud to something like 17,000' or more AGL.  Again he had a turn rate indicator but no artificial horizon.

@JohnK -- see also https://aviation.stackexchange.com/a/48150/34686 (last paragraph), https://aviation.stackexchange.com/a/26811/34686 , https://aviation.stackexchange.com/a/63971/34686, https://aviation.stackexchange.com/a/63974/34686 . See also comment by yankeekilo https://aviation.stackexchange.com/q/26808/34686 .

@JohnK -- Also worth noting that cloud flying used to be allowed in sailplane competition, but sometime (perhaps in the 50's or 60's?) that came to be banned in contests in most countries, and by now that is universally the case as far as I know. One obvious reason is the collision hazard. Turn rate indicators (and all other gyro instruments) have been banned in gliders during contests in most countries for decades, because it is obvious that such an instrument would facilitate violating the contest rules (and w/ modern airspace structures, violating FARs etc) and getting an unfair advantage

Some material in the above comments would probably make a useful addition to some of the linked answers; I've saved to clipboard for future use--

@quietflyer You are clearly abusing the comment feature for things it is not meant to do. Why don't you ask a new question and add all that material to the answer? But keep in mind that VFR cloud distances must be observed by gliders, too.

@PeterKampf --  Brings up the can-of-worms question as to what you are allowed to do in uncontrolled airspace without filing an IFR flight plan, if you possess an instrument rating or other appropriate rating (I believe a glider cloud flying rating exists in the UK).   In the UK I believe the uncontrolled airspace often goes higher than the 700' or 1200' ceiling we usually have in the US.

@PeterKämpf -- In the one case I mentioned in the US, I don't know what was the floor of the controlled airspace at that time, or whether the pilot was actually flying under IFR rather than VFR-- discussed more here https://aviation.stackexchange.com/q/78746/34686

Lots of good information in answers posted so far but I think it is also useful to point out that with no static stability (in the pitch axis), the aircraft wouldn't be trimmable.  With positive static stability (in the pitch axis), you can trim for a given airspeed, and if you then pull the stick aft, you'll feel an increasing forward pressure on the stick, and if you then release the stick, the nose will drop and the airspeed will increase.  Likewise, if you trim for a given airspeed and then push the stick forward, you'll feel an increasing aft pressure on the stick, and if you then release the stick, the nose will rise and the airspeed will decrease.  Static stability makes that possible.
But that's not sufficient to ensure that the aircraft will actually settle down at the trim airspeed after a disturbance, if you continue to exert no forward or aft pressure on the stick or yoke.  As other answers have noted, if the aircraft is statically stable but dynamically unstable (in the pitch axis), after an initial disturbance, it can enter a pitch "phugoid" oscillation that keeps getting bigger, rather than damping out, unless the pilot intervenes.  Here's a video showing such a thing.
It's worth noting that these effect can be subtle enough that a pilot can accumulate many hours of hands-on flight time in some particular aircraft, without actually having any idea whether it is dynamically stable or unstable in the hands-off case, in any given configuration.

why does that real airplane not have dynamic stability? I thought engineers would design an airplane that has ocsillation damped out?

@Dat -- it's a glider, and see the reference to gliders in other answer here https://aviation.stackexchange.com/a/83369/34686

Static stability is the initial response of a plane to an instant impulse (like a turbulence), while dynamic stability is how a plane responds over time to a disturbance.
Dynamic stability can be verified by pulling/pushing one flight control surface and instantly letting it go: oscillations on the related axis can increase in amplitude, decrease in amplitude or maintain the same amplitude.
If, over time, the plane is stable then, the initial response is stable, but you can't say the opposite.
I don't know to explain it mathematically (if it's what you are looking for).
References: https://www.boldmethod.com/learn-to-fly/aerodynamics/3-types-of-static-and-dynamic-stability-in-aircraft/


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How do I calculate the total cruise power from the propellers accounting for the losses?
Let's say I have an aircraft with four propellers (see diagram below) and I am supposed to calculate the cruise power required to drive the aircraft forward. The  wing creates the lift and propellers provide the forward thrust.
If I don't account for the wake of the front propellers affecting the rear propellers it's easy. I just have to calculate the thrust required and get the power. But how do I account for the losses and the actual power required?

Are the four rotors only providing forward thrust and the wing provides the lift?

Yes ,Wing provides the lift and the rotors provide the forward thrust

Why is there downwash? What angle are the rotors?

90 deg. its not the downwash I meant to say the flow leaving the front rotors

If you counter rotate them the losses might be gains.

How is it possible?
The wake from the front rotors will be highly turbulent right

@sai teja -- I think highly turbulent left for the left one.

The flow leaving the propeller:

Contracts, as a function of distance behind the propeller. The impulse theory gives a contraction factor - for an idealised homogeneous outflow field. The contracted wake has a higher velocity than free stream.
Has a whirl component due to the drag of the blades.
Has propeller tip vortices in a contracting corkscrew field.

So if the aft propellers are far enough behind the forward ones, the aft propeller tips are unaffected by the wake of the forward ones.
If the aft prop counter-rotates, the whirl component from the front props will be removed. Optimal blade angle of the aft prop is different from that of the fwd prop, but the tips would remain relatively unaffected.

Picture above is from Leishman, which has a whole chapter devoted to rotor wake. There has been quite some research done on co-axial rotors for helicopters. There is some info and a link in this answer as well.
But I reckon you'll be in the right order of magnitude if you just compute required power as you would in an undisturbed flow, especially if you use ducted propellers which reduce the tip vortices.

I appreciate your answer but can you tell me in terms of percentage how much will be the loss?..Can it be neglected

The Ray Prouty book mentions about co-axial rotors: "Separate them 20% of the rotor radius, and you still have a 7% power advantage over a single 4-blade propeller, as computed by Kamov [(reference)][3]." Counter rotating props can actually have a power advantage - a gain, not a loss. That would mean that losses can be neglected.


text:
How do uncoordinated turns (skid, slip) affect a helicopter's turn radius?
Is a helicopter's flight path (track made good) an arc of constant radius regardless if it's a coordinated or uncoordinated turn? (Assume nil wind conditions and that the pilot maintains the angle of bank, altitude and airspeed constant in both cases.)
I'm confused because texts mention slipping and skidding as sliding downhill and skidding uphill from the turn respectively, which to me reads like the radius of the turn changes.
[EDIT] Updated assumptions to include bank angle and airspeed.

"Assuming a constant altitude" is a big assumption. In a turn, power must be added to maintain acceleration. This in turn increases torque, and therefore all kinds of side-effects. I've consulted Prouty and Wagtendonk (perhaps the two leading authorities on rotary dynamics) and I can't find an answer to this. My intuition says that yes, the radius will be constant since with no wind, there is nothing to disturb the equilibrium except, perhaps, for the helicopter encountering it's own vortex (which is a good test of skill when making constant attitude coordinated turns) but I really don't know.

Given your assumptions about bank angle and airspeed and my assumption that coordinated turn is determined by acceleration inside aircraft pointing downward "perpendicular to the floor" (if you mean anything else by coordinated turn, try to be more specific on it), then you have one exact radius for coordinated turn. That one where centrifugal force generated by airspeed and curvature summed with gravity gives vector matching the (fixed) bank angle. If you want local acceleration to point in other direction (making slip or skid), only free variable remaining is turn radius.
So yes, given airspeed and bank angle is constant, turn radius is the parameter determining if turn is coordinated or not.
In a slip the local acceleration in aircraft points toward that side you want to turn (or, in other words, aircraft's floor is banked too much outside the turn). To get this, centrifugal force has to be less compared to coordinated turn (centrifugal force is only quality you allowed to change), so turn radius have to be larger. To get a skid turn radius have to be smaller for the same reason.
Inertial and gravity acceleration as seen and felt inside an aircraft during a turn with bank angle $\alpha$:

where $\vec{g}$ gravity acceleration and $\vec{a_c}$ centrifugal acceleration. Resulting $\vec{a}$ is total acceleration felt by pilot. From basic mechanics $a_c=v^2/r$ where $v$ and $r$ is speed (either TAS or GS, both are the same here) and turn radius respectively. Turn radius for coordinated turn is $\tan\alpha=a_c/g\Rightarrow r_{\rm coord}=v^2/(g\tan\alpha)$. Any smaller radius with the same bank and speed leads to higher $a_c$ and thus skid, larger radius results in slip. Note that this is true for any aircraft, not specific for helicopter.

Martin, given that you've stated the turn radius determines whether its a coordinated turn (as we've assumed other parameters remain fixed in this example), would the respective flight paths then look like ?

@jumblie I believe, it is just other way around. Slipping turn gives (with the same bank angle) larger turn radius than coordinated one, skidding turn smaller. If it is not clear from images above, you can try to imagine it like this: would you fly according to the "step on the ball" rule, then in slipping turn the ball is "inside turn" and you have to step on bottom ruder to get coordinated turn. That is you need to "turn more".

One could tend to believe slip gives "more turning" because we have sideslips, right? But note that sideslip (or forward-slip, it is the same maneuver aerodynamically anyway, only ground track differs) is no turning at all, it is airplane flying straight forward, only difference is the nose points "wrong way".

I've amended my diagram so that the tighter turn is the skidding turn and the larger turn is the slipping turn, but its still not quite clear in my head. Regardless if the turn is a coordinated/slipping/skidding one, will the aircraft always be tangential to the track? Also, sliding downhill makes it sound as if the radius would get smaller and skidding uphill as if the radius would get larger but its exactly the other way around!

@jumblie IMO has any aircraft in skid/slip to fly partially sideways. Approximately like you have sketched it in your drawing. Reason becomes clear when you look on forces from outside. You need, in addition to the lift some aerodynamic force pushing aircraft to the side. Reasoning will be little different for fixed-wing and helicopter (because rotor teeter allowing lift to point somewhat  aslant relative to helicopter's fuselage -- someone actually flying helicopter can correct me here if I am wrong), but the result should be same. I'll try to add some explanation into my answer eventually.

@jumblie I am not sure where does the up/downhill simile come from, so I am not able to elaborate on this too much. Maybe it has something to do with what pilot feels during skidding/slipping turn? See images in my answer. In the slipping turn resulting force felt in cabin pulls you toward turn center, which can be seen as "downhill" if you think angle of your seat relative to the horizon. During skidding turn net force points partially outwards, which could be imagined, analogically, as uphill? Could it be so?

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why does an aircraft pitch up when the speed increases?
Can anyone please explain why aircraft pitches up when the speed increased? (please consider an aircraft, Wing AC, CG and Tail AC lies on a line.)
The ideal explanation I'm looking for should have something to do with the static margin.
Here I am talking about a powered or unpowered aircraft, trimmed for steady flight, reacting to an EXTERNAL velocity perturbation which increases the airspeed. (Like a sudden, substantial, sustained increase or decrease in the speed of the wind, e.g. due to wind shear.)
I have seen that when faced by this kind of a perturbation, the aircraft tries to slow down/keep the airspeed unchanged by increasing the pitch angle.  My question is how it happens
Peter Kampf wrote:

However, I can imagine what might have happened to let you observe a pitch-up. This needs several conditions:
      A propeller-diven airplane with the propeller in the front
      A sufficiently large static margin so the empennage produces a down-force.
      Speeding up by opening the throttle.

Let me clarify these are not necessarily the cases for what I'm asking,
    1. If we throw a glider hard enough, which is trimmed for a certain airspeed it will pitch up and climb.
    2. At least for now I think it doesn't matter whether the tail produces down-force or upward force. Say the velocity is increased by factor of 2 and the forces on both wing and tail increased by a factor of 4. Equilibrium is still maintained as moment around CG has not changed by this. (and this is what stands against my observation)
Mike Sowsun wrote:

The horizontal stabilizer always provides a downward force to balance the forces of lift and weight with the center of gravity. This also provides stability because if the aircraft pitches down and starts to speed up, the increased airflow over the tail will result in more downward force and cause the nose to rise and the aircraft to slow.

I think there is no need for the tail to make downward lift always. The tail can be uplifting as well. Anyway, if we agree that to be the case for now, when the aircraft gains airspeed, the flow speed over the wings increases as well as tails. Isn't it? What I can't understand is, what makes the aircraft pitch up when both the wing force and tail force increased by the same factor, due to increase of airspeed.
Any further input is much appreciated.

I am not familiar with the term 'static margin', but a partial answer for the nose pitch up when airspeed increases is due to the increase in the downward or 'negative' lift over the horizontal stabilizer/elevator due to the higher airspeed. This causes a pivoting action around the center of pressure on the wing thus increasing the angle of attack and moving the CP forward, often, away from the CG.     I know it's not the full answer but hopefully this will help.

You are wrong. The tail must ALWAYS provide a down force in a conventional aircraft. That is why increased airspeed causes a pitch up.

@MikeSowsun: May I respectfully disagree? All what is needed for stability is less lift per area on the tail than on the wing.

Yes, I suppose that is true for some a aircraft, in some conditions. The principle remains the same though. Faster speed means that the main wing provides more lift than the tail and the nose will pitch up. Still, everything I have read says conventional aircraft tails provide a down force.

You should read how it flies?

@MikeSowsun: The OP specifically mentions gliders that he can throw with his hands. My experience with free-flight model airplanes, especially indoor competition models, is that they tend to use lifting tails. Static margin for these planes are crazy. Most of my models have CG at 60-80% MAC. I even have planes with CG behind the trailing edge of the main wing.

No really good answers have been offered to this question yet!

This question needs to be edited to remove the idea that turning into a headwind will tend to make the airspeed increase.

@quietflyer: How would an increasing headwind not make one's airspeed increase (at least initially)?

@Sean - That's pretty basic. From the airmass reference frame the wind doesn't exist, except as an apparent motion of the ground. From the ground reference frame it at first glance seems paradoxical to explain what exactly is the source of the force that causes the change in groundspeed as a plane circles but it can be shown that it is simply the lift force from the banked wing, which does not always act purely perpendicular (centripetal) to the ground track, so is capable of causing a longitudinal acceration as well as a centripetal one.

@Sean - But best to stick with the airmass reference frame till you are comfortable with the idea that turning into the wind does not make the airspeed increase.  The plane doesn't "know" the wind is there.

"Let me clarify these are not necessarily the cases for what I'm asking, 1. If we throw a glider hard enough, which is trimmed for a certain airspeed it will pitch up and climb." -- I would suggest that this is actually exactly what the question is asking, in essence.

What you ask is: If a model glider is thrown at a higher speed than its trim speed, why does it pitch up?
Short answer: Because the rear horizontal surface produces less lift per area than the forward surface. When flying at a speed different from its trim speed, the combined center of lift of all surfaces is shifted such that it creates a pitching moment around the center of gravity. This pitching moment causes the flight path change.
Note that I try to explain things in terms that work equally well for canards or even flying wings. That may sound odd in places, but needs only one explanation for all cases. For unswept flying wings read forward wing = forward part of the airfoil and vice versa.
Long answer: Let's assume an airplane that is rigged for flight at an angle of attack of 9°. For simplicity, let's assume that both airfoils are symmetric and both surfaces have the same lift curve slope, for simplicity 0.1 per degree. The local incidence is 0° on the forward and -5° on the rear wing. Neglecting downwash effects, this results in a lift coefficient on the forward wing of 0.9 and on the rear wing of 0.4. The zero-lift angle of attack is +1° when the forward wing lift coefficient is at 0.1 and the rear wing lift coefficient is at -0.4.
Let's further assume that the tail surface has 25% of the area of the forward surface. Lift on the forward wing is 90% of weight and on the rear wing it is 10% of weight. The center of lift, therefore, sits at 10% of the line connecting both neutral points and that is where the center of gravity is as well for trimmed flight. Like this:

Now the model is thrown at twice its trimmed speed. Let's assume that the angle of attack is also 9°, but that does not matter much. Lift on the forward surface is now 360% and on the rear surface 40% of weight. Again both forces combine for a common center of lift at 10%, so no pitching moment is created. But lift far exceeds weight, so the aircraft immediately climbs. Climb without pitching motion means that the angle of attack decreases immediately. Therefore, the aircraft will slightly accelerate upwards and settle at a new, lower angle of attack where the combined lift is equal to weight. But how is lift now distributed?
To have lift decrease to a quarter needs an angle of attack change to 25% of its initial value, relative to the zero-lift angle of attack. For this, the new angle of attack on both surfaces has to decrease by 6°. The local incidences result in an angle of attack of 3° on the forward and -2° on the rear wing and lift coefficients of 0.3 and -0.2, respectively. Again neglecting downwash effects, the new lift contributions are 120% on the forward wing and -20% on the rear wing. Like this:

Now the center of lift is 20% of the distance between the two wing's neutral points ahead of the forward wing's neutral point and 30% ahead of the center of gravity. That causes a strong nose-up pitching moment which will let the aircraft pitch up. Together with the initial flight path change of 6° for lift correction this will make the airplane climb until its flight speed drops below the trimmed condition and the condition reverses. Since pitch damping is high, very few cycles are needed to arrive at the trimmed condition but at a height over the launch point which corresponds to the extra energy provided by the high launch speed.
This can be run with different numbers and will work regardless of lift or downforce on the rear surface, provided that lift per area is lower there than on the forward wing.

@ PeterKampf, How does turning into a headwind cause airspeed to increase?

@PeterKampf-- "In a glider this happens only when turning into a headwind - the true airspeed increases and momentarily increases the flight path angle which looks like pitching up for the pilot." -- surely not.  Circling in wind so strong that I'm drifting backward when pointing straight into the wind, I never experience this.  This answer could be improved by deleting this sentence.  This incorrect idea is the basis of a faulty theory of dynamic soaring that someone has been propagating lately, in contrast to the correct theory of dynamic soaring based on the wind gradient.

@PeterKampf -- also, why would you need to specify a front-mounted propeller, in order for opening the throttle to cause a pitch-up?  I would suggest that pitching up and climbing is the normal response to an increase thrust in almost any aircraft.  Any pitch-down tendency due to a decrease in angle-of-attack would be dwarfed by a pitch-up tendency due to the increased climb angle.

"Climb without pitching motion means that the angle of attack decreases immediately. Therefore, the aircraft will slightly accelerate upwards and settle at a new, lower angle of attack where the combined lift is equal to weight. Now vertical acceleration has stopped. But how is lift now distributed?" -- this model seems to break things down into discrete steps that may not actually exist-- do you really believe the vertical acceleration falls to zero at this point shortly after the plane starts to climb?

Consider the case where the initial excess airspeed is so extreme that the plane ends up flying a full loop.  Still, the approach does provide insight into why the a/c tends to maintain the trimmed angle-of-attack regardless of airspeed..

@quietflyer: I repeatedly mentioned simplifications. This is one more. Of course things are happening together and not in stages. But by dissecting the complex whole in simpler parts should aid understanding.

The horizontal stabilizer always provides a downward force to balance the forces of lift and weight with the centre of gravity. This also provides stability because if the aircraft pitches down and starts to speed up, the increased airflow over the tail will result in more downward force and cause the nose to rise and the aircraft to slow.
As the aircraft further slows the decreased airflow over the tail will cause the nose to drop and the airspeed to increase again.
This pattern will then continue in a "Phugoid  motion"

The horizontal stabilizer produces less lift than the wing but not necessarily negative lift. And your answer does not take into account canard configuration (that works in a similar way)

The horizontal stabilizer need not always produce a downforce.  Many free-flight model airplanes are set up with lifting horizontal stabilizers.

"This also provides stability because if the aircraft pitches down and starts to speed up, the increased airflow over the tail will result in more downward force and cause the nose to rise and the aircraft to slow" -- the problem with this line of argument is that it tends to suggest that if you add weight to the CG of a glider, so that it tends to fly faster, it will have some tendency to trim to a higher angle-of-attack, which isn't true.

The real mechanism behind speed stability is more complicated and includes the fact that if the flight path tends to curve upward for any reason, gravity has now gained a component that acts parallel to the drag vector, so the aircraft will tend to slow.

(Yes I see the thread is some years old.)

The aircraft will stay at the trimmed angle of attack with an increase in airspeed but will accelerate upward due to the increase in lift,which is dependent on airspeed (there is no pitch-up, which is a misnomer for this case- pitch-up being defined as an increase in AoA). As the craft rises up, it continues in a constant AoA curve until the longitudinal speed decays to zero, at which point it falls down of course- causing an abrupt change in the wing angle of attack which results in a stall. As the craft tries to regain the trim AoA due to its inherent stability, its motion becomes a series of climb, stall, and descent oscillations, due to attempts to damp out AoA overshoots in the process.  Note that with sufficient initial launch speed it will continue in its constant AoA curve in a loop which will continue on the reverse side until its trim airspeed is once again regained- without any series of climbs and stalls-since the AoA was maintained in the first place.
The explanation here is for a free-flight model with preset control surfaces,
no power or constant- thrust power.

Why is a pitch-up "defined" as an increase in angle-of-attack?  An increase in thrust will tend to cause an aircraft to pitch up and climb.  Yes, the short-term dynamics involved in making this happen do involve a temporary increase in lift due to a temporary increase in airspeed, which may be almost imperceptible to the pilot.

I think I've got the answer. The key is, When the speed increased, it appears to the aircraft that the angle of attack has decreased.
Once the angle of attack decreased, the rest works exactly as normal longitudinal stability case.
http://adg.stanford.edu/aa241/stability/staticstability.html

please consider stop making new accounts, if you register you will be able to edit you own posts. and also do not post comments as answers

I'd suggest that an increase in airspeed looks more like an increase in angle-of-attack, not a decrease in angle-of-attack, to the airplane.  Both an increase in airspeed and an increase in angle-of-attack cause the lift vector to increase which causes the flight path to curve upward into a climb.  Ultimately of course the aircraft will stabilize in a steady climb with SMALLER lift vector than it had in level flight.

You ask "What I can't understand is, what makes the aircraft pitch up when both the wing force and tail force increased by the same factor, due to increase of airspeed."
It is critical to understand that an imbalance in pitch moment-arm between the wing and the tail is only required to cause a CHANGE in the rate of pitch rotation, not to cause a pitch rotation.  In general, you aren't going to be able to account for pitch rotation by looking for an imbalance in pitch moment-arm between the wing and the tail.  Any explanation that involves an imbalance in pitch moment-arm between the wing and the tail, is getting down into the nitty-gritty details about the cause of a CHANGE in the pitch rotation rate, and how this CHANGE in pitch rotation rate then restores the balance in pitch moment-arm between the wing and tail.
Taking a larger view, it may be sufficient to simply understand that to a first approximation, an aircraft tends to trim to a constant angle-of-attack, because any deviation from the trim angle-of-attack tends to set up an imbalance in pitch moment-arm between the wing and the tail, which creates a pitch torque that changes the pitch rotation rate, which leads to a change in angle-of-attack.  (Later we'll see why this statement is only an approximation-- why an aircraft deviates somewhat from the trim angle-of-attack during a pitch "phugoid" oscillation.)
What you are asking about is called "speed stability".
Many attempted explanations of this phenomenon suffer from the following flaw-- they suggest that if we add weight at the CG of a glider, this will cause the glider to trim not only to a higher airspeed, but also to a different angle-of-attack.  This is not accurate.
Another common flaw in explanations of this phenomenon is an implicit assumption that the flight path will initially remain horizontal after a sudden increase in airspeed (which implies that that the angle-of-attack must be well below the trim angle-of-attack), until the aircraft starts to pitch up.
Let's see if we can offer an explanation of "speed stability" that does not suffer from either of these flaws.
Imagine the aircraft is flying into the wind, and the wind suddenly increases by 20 mph.
Ultimately, the plane may return to equilibrium at its original airspeed and a lower groundspeed.
However, what happens in the short term due to the instantaneous 20 mph increase in airspeed?
To a first approximation, the aircraft tends to maintain its trimmed angle-of-attack.  Drag is greater than thrust, so airspeed is decreasing, but is still higher than the trimmed airspeed.
The excess airspeed is creating excess lift, so the flight path starts to curve upward.  The excess lift is acting as a "centripetal force" and forcing the flight path to curve.
As the aircraft tends to maintain its trimmed angle-of-attack and the flight path starts to curve upward, the nose must rise.
This could be the end of the answer right here.  But in case you want to know what happens next...
As flight path curves upward, gravity gains a component acting parallel to the drag vector and against the thrust vector, further contributing to the rate of decrease in airspeed.
At some point in the climb, as the airspeed continues to decrease, causing the lift vector to continue to decrease, the force acting "upward" relative to the flight path (i.e. the lift vector) becomes less than the force acting "downward" relative to the flight path (i.e. a component of the weight vector).  At this moment the flight path stops curving upward and starts curving downward.  The nose starts to fall back toward the horizon, and then falls below the horizon, even as the aircraft is still approximately maintaining its trimmed angle-of-attack.
When the flight path curves below horizontal, gravity gains a component acting parallel to the thrust vector and against the drag vector.  Shortly before this point is reached, the balance of fore-and-aft forces in the aircraft's reference frame must be such that the airspeed will start increasing again.
Eventually the rising airspeed will increase the lift vector to the point where the flight path stops curving downward and starts curving upward again, as the cycle continues. The whole cycle is known as a pitch "phugoid" oscillation.
The complete process of returning to equilibrium may involve several, or many, slowly-decreasing cycles of the pitch "phugoid" oscillation.
A complete explanation must recognize that in actual practice the angle-of-attack does not stay absolutely constant throughout the pitch "phugoid", both due to rotational inertia in the pitch axis, and due to aerodynamic effects caused by the curving flight path and the resulting curvature in the undisturbed relative wind, or to put it another way, aerodynamic damping in the pitch axis.  The angle-of-attack tends to be highest near each peak in altitude, and tends to be lowest near each low point in altitude, due to the direction that the flight path is curving at each of these points.  (In some aircraft this can be demonstrated by setting up the initial conditions so that the stall horn sounds near each of the altitude peaks, even as the pilot keeps his hands off the controls.)  This is not the fundamental driver of the pitch phugoid; in many cases we'd still see a rather similar pitch phugoid even if we manipulated the controls to keep the angle-of-attack exactly constant.  However in extreme cases, we can get a full-blown stall break near each of the altitude peaks, in which case the oscillation is unlikely to damp out.
And here's an even more extreme case to consider: if we trim an aircraft for level flight at 50 knots, and then we dive to 100 knots and then quickly pull out into level flight and release controls to allow the aircraft to return to the trim angle-of-attack, the situation may be so extreme that the flight path may curve well past vertical into semi-inverted flight, or may even describe a complete loop, or the aircraft may run out of airspeed going nearly straight up and violently "whip stall".  So there's a limit to the situations that we can expect will lead to a gentle, tame, pitch "phugoid" that smoothly damps out.
You may be hoping for more detail on exactly why the airplane tends to maintain its trimmed angle-of-attack as the airspeed varies, and also why in actual practice we do see a some deviation from this principle as described above.  These are not a simple topics.  A good explanation of the basic tendency to maintain a trimmed angle-of-attack may be found in this section of John S. Denker's excellent "See How It Flies" website-- "6  Angle of Attack Stability, Trim, and Spiral Dives" -- https://www.av8n.com/how/htm/aoastab.html .  Decalage is key, but the horizontal stabilizer need not actually create a downforce.
Propwash is another factor that complicates (interferes with) an aircraft's tendency to trim to the same angle-of-attack regardless of airspeed; nonetheless most single-engined powered planes with tractor propellers, jets, gliders, and aircraft of other configurations all behave in GENERALLY the same way, and examples of the dynamics described in this answer can be observed in all of them.

I should have written this answer in two-part form, "short answer" followed by "longer answer".   The "short answer" is simply that to a first approximation the angle-of-attack stays constant, but any excess airspeed creates excess lift which will curve the flight path upwards, so the nose rises.  Exactly as happens if you trim for hands-off pitch control in a steep turn, and then roll abruptly to wings-level.  To a first approximation the angle-of-attack stays constant, (ctd)

(ctd) but the airspeed (and therefore the lift force) temporarily retained from the steep turn is more than is needed for linear wings-level flight, so the flight path curves upwards, taking the nose along with it.  In extreme cases (easy provable with radio-controlled models, if you don't wish to attempt it in a real life) you can trim for constant-airspeed flight in a steep bank and then roll to wings-level so quickly that the airplane or glider will pull up into a full-blown loop due to the dynamics described above.

I think it has a very simple logic As static margin is the distance between the center of gravity and the neutral point of the aircraft.Whenever thrust is increased the wings of aircraft provide an upward thrust due to the design and pressure difference.This thrust is then accompanied by the static margin affect keeping an unbalancing force on the rear portion resulting into Pitch moment.

authors: Людмил Станчев

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/